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Task

Write a program/function that when given a positive integer \$n\$ splits the numbers from \$1\$ to \$n\$ into two sets, so that no integers \$a, b, c\$, satisfying \$a^2 + b^2 = c^2\$ are all in the same set. For example, if \$3\$ and \$4\$ are in the first set, then \$5\$ must be in the second set since \$3^2+4^2=5^2\$.

Acceptable Output Formats:

  • One of the sets
  • Both the sets
  • An array of length \$n\$ where the \$i\$-th element (counting from 1) is one of two different symbols (e.g. 0 and 1, a and b, etc.) which represent which set \$i\$ belongs to.The reverse of this is also fine

Constraints

You can expect \$n\$ to be less than \$7825\$. This is because \$7824\$ is proven to be the largest number to have solution (which also implies that all numbers less than 7825 have a solution).

Scoring

This is so shortest bytes wins.

Sample Testcases

3 -> {1}
3 -> {}
5 -> {1, 2, 3}
5 -> {1, 2, 3}, {4, 5}
5 -> [0, 0, 0, 1, 1]
5 -> [1, 1, 0, 0, 1]
10 -> {1, 3, 6}
10 -> {1, 2, 3, 4, 6, 9}
41 -> {5, 6, 9, 15, 16, 20, 24, 35}

A checker to verify your output can be found here

Inspired by The Problem with 7825 - Numberphile

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9
  • 1
    \$\begingroup\$ Is it guaranteed that a solution exists for all n below 7825? It may be worth clarifying in the challenge \$\endgroup\$ – Luis Mendo Jun 20 '20 at 16:04
  • \$\begingroup\$ Suggested test case: 41. This is the first value for which this simple but invalid algorithm doesn't work anymore: start with an empty list A; for x = 1 to n: if there's some x in A such that sqrt(x²+n²) also exists in A: leave A unchanged else append x to A; return A. \$\endgroup\$ – Arnauld Jun 20 '20 at 16:08
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    \$\begingroup\$ @LuisMendo Theorem 1. The set {1, . . . , 7824} can be partitioned into two parts, such that no part contains a Pythagorean triple, while this is impossible for {1, . . . , 7825}. arxiv.org/pdf/1605.00723.pdf \$\endgroup\$ – ZaMoC Jun 20 '20 at 16:33
  • \$\begingroup\$ @Arnauld added 41 as a testcase \$\endgroup\$ – Mukundan314 Jun 20 '20 at 16:52
  • \$\begingroup\$ May we output an empty set when \$n<5\$? (I would assume so.) \$\endgroup\$ – Jonathan Allan Jun 20 '20 at 19:25
7
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J, 37 bytes

Brute forces through the possible sets, outputs the bit mask.

((-&.#.+./@,)[(e.~+/~)/.*:@#\)^:_@#&1

Try it online! (Also outputs list as numbers for easier comparison.)

How it works

((-&.#.+./@,)[(e.~+/~)/.*:@#\)^:_@#&1
                                  #&1 convert to list of N 1's
(                            )^:_     do until list does not change
                        *:@#\         right: convert to 1,4,9…,N^2
             [                        left: the bit mask
                      /.              partition left based on right, for each set:
                  +/~                 make M*M addition table
               e.~                    any element of that in the same set?
       +./@,                          OR all answers: 1 on conflict, 0 if finished
  -&.#.                               list: from base 2, subtract that^, to base 2
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2
  • \$\begingroup\$ I guess 1,2,4…,N^2 is a typo of 1,4,9…,N^2? \$\endgroup\$ – Bubbler Jun 21 '20 at 23:27
  • \$\begingroup\$ @Bubbler Indeed! \$\endgroup\$ – xash Jun 21 '20 at 23:39
6
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Jelly, 18 bytes

œc3²SHeƊ$Ƈ
ÇŒpÇÞḢQ

Try it online! (too inefficient for \$n>25\$ on TIO).

How?

Strategy: Find all Pythagorean triples using \$[1,n]\$ then find a way to pick 1 element from each of them such that the resulting set contains no Pythagorean triples. That way we have a set which both contains no Pythagorean triple and blocks the other set from having any.

œc3²SHeƊ$Ƈ - Link 1, find all Pythagorean triples: list of integers OR number
œc3        - all combination of length 3 (given n uses [1..n])
         Ƈ - keep those for which:
        $  -   last two links as a monad:
   ²       -     square each of them
       Ɗ   -     last three links as a monad:
    S      -       sum (of the three squares)
     H     -       halved
      e    -       exists in (the squares)?

ÇŒpÇÞḢQ - Main Link: n
Ç       - call Link 1 as a monad -> all Pythagorean triples using [1,n]
 Œp     - Cartesian product -> all ways to pick one from each
    Þ   - sort those by:
   Ç    -   call Link 1 as a monad (empty lists are less than non-empty ones)
     Ḣ  - head
      Q - deduplicate (if n < 7825 this is a valid answer)
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11
  • 1
    \$\begingroup\$ Nice! Certainly runs faster than mine, besides being much shorter \$\endgroup\$ – fireflame241 Jun 20 '20 at 19:30
  • \$\begingroup\$ What is the purpose of ÇÞ? Isn't the Cartesian product automatically in that order? \$\endgroup\$ – fireflame241 Jun 20 '20 at 19:40
  • 2
    \$\begingroup\$ No, once \$n=26\$ the result's first item will contain \$6\$, \$8\$, and \$10\$. (So taking the smallest of each triple is not a valid strategy.) \$\endgroup\$ – Jonathan Allan Jun 20 '20 at 19:45
  • \$\begingroup\$ Oh, I misunderstood how Þ works, and I was about to bring up the counterexample for n=86 ((5,12,13),(12,35,37),(13,84,85)) \$\endgroup\$ – fireflame241 Jun 20 '20 at 19:47
  • 1
    \$\begingroup\$ In that case, I retract my objection - assuming that, as you pick more than one due to repeats, you check that you don't end up picking all of a triple eventually. If you have a valid partition into two sets X and Y, you can just, for each triple, pick an element of that triple that's in X. The set Z of all elements picked is a subset of X (so it doesn't contain all of any triple) but contains one element of each triple by construction. \$\endgroup\$ – Misha Lavrov Jun 21 '20 at 20:13
5
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Wolfram Language (Mathematica), 132 116 bytes

{1}.SatisfiabilityInstances[And@@(And[Or@@#,Nand@@#]&/@Map[x,Select[#~Tuples~3,{1,1,-1}.#^2==0&],{2}]),x/@#]&@*Range

Try it online!

This uses Mathematica's SAT solver to label the integers 1 through the input as True and False.

  • This is composed with Range, so what feeds into the main function is a list of the integers from 1 to the input.
  • Select[#~Tuples~3,{1,1,-1}.#^2==0&] generates all the Pythagorean triples (multiple times actually, but that's okay).
  • And[Or@@#,Nand@@#]& is true if at least one, but not all, of the elements of its input is true.
  • {1}.SatisfiabilityInstances[...,x/@#] uses the SAT solver. Since SatisfiabilityInstances returns a list containing one solution, we use {1}. to get its first element.
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4
  • \$\begingroup\$ Very nice +1! you can use infix notation in ~Subsets~ to save a byte. Also your program must be a function and that costs 3 bytes... Try it online! \$\endgroup\$ – ZaMoC Jun 20 '20 at 21:35
  • \$\begingroup\$ @J42161217 My program is a function, I just haven't finished it with &: it's the composition of two functions, {1}.Satisf....,x/@#]& and Range. Your version composes with Range@#& instead which is equivalent to Range, but longer. \$\endgroup\$ – Misha Lavrov Jun 20 '20 at 21:40
  • \$\begingroup\$ My bad, here is a better presentation 131 bytes! \$\endgroup\$ – ZaMoC Jun 20 '20 at 21:57
  • \$\begingroup\$ I ended up going with Tuples instead, because #~Tuples~3 is shorter than #~Subsets~{3}. It also generates way more triples, and even after the Select we end up getting {3,4,5} and {4,3,5} separately, but that's fine. \$\endgroup\$ – Misha Lavrov Jun 20 '20 at 22:07
5
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JavaScript (ES6),  118  117 bytes

Much slower for -1 byte.

f=(n,a=[],b=a)=>n?f(n-1,[n,...a],b)||f(n-1,a,[n,...b]):[a,b][E='every'](o=>o[E](x=>o[E](y=>o[E](k=>k*k-x*x+y*y))))&&b

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JavaScript (ES6),  122 119  118 bytes

Returns one of the sets as an array.

f=(n,a=[],b=a)=>[a,b][S='some'](o=>o[S](x=>o[S](y=>o[S](k=>k*k==x*x+y*y))))?0:n?f(n-1,[n,...a],b)||f(n-1,a,[n,...b]):b

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Solution found locally for \$n=41\$:

[ 5, 6, 9, 15, 16, 20, 24, 35 ]
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4
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05AB1E, 14 bytes

Port of the 17 byte Jelly answer. (Læ3ùʒDnO;tå}€н is the same length)

Læ3ùʒnRćsOQ}€н

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Explanation

L              Length range
 æ             Powerset
  3ù           Pick truples (length-3 tuples)
    ʒ          Filter:
     n             Square all items
      R            Reverse the list
       ć           Head-extract (head on top)
        s          Swap
         O         Sum the remaining list
          Q}       Equal?
            €н Take head of each
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1
  • \$\begingroup\$ Not sure how to fix it, but it's currently incorrect for certain inputs \$\geq26\$, just like Jonathan's initial answer. PS: using 3.Æ is faster than æ3ù, although it doesn't matter for the byte-count. :) \$\endgroup\$ – Kevin Cruijssen Jun 22 '20 at 7:21
3
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Jelly, 30 26 bytes

œ|/L=³
Œc§œ&
ŒP²ÇẸƊÐḟŒcÑƇḢ

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How?

This does a more brute-force approach, filtering subsets of [1..n] based on whether they contain any Pythagorean triples. Then, it finds two triple-less subsets that have all n elements between them

œ|/L=³         # Test if a pair of sets unions to [1..n]
œ|/              # Set intersection  
   L             # Is the length
    =³           # equal to n?       

Œc§œ&          # Does a pair exist that sums to another?
Œc               # Compute all pairs of squares
  §              # Sum each
   œ&            # Set intersection with the set of squares (nonempty & truthy if a pair of squares sum to another square)

ŒP²ÇẸƊÐḟŒcÑƇḢ  # Main link
ŒP               # All subsets of 1..n
     ƊÐḟ         # Remove those where:
  ²                # of the squares,
   ÇẸ              # a pair of the squares exists that sum to another square
        Œc       # All pairs of these triple-less subsets
          ÑƇ     # Filter the pairs by whether they union to [1..n]
            Ḣ    # Head; get the first one
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2
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Wolfram Language (Mathematica), 1664 bytes

works for all n (1 to 7824) instantly

IntegerDigits[Uncompress@"1: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",2][[;;#]]&

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2
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R, 99 95 bytes

n=scan():1
f=function(j)outer(a<-n[j]^2,a,`+`)%in%a
while(any(f(i<-sample(!0:1,n,T)),f(!i)))0 
i

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Outputs a vector of TRUE and FALSE representing in reverse order which set each integer belongs to. (The footer of the TIO transforms this into a list of integers in the first set.)

Works by random sampling: repeatedly draw a random subset of 1:n until neither the subset nor its complement contain any Pythagorean triples (checked by the function f).

It will finish in finite time for any input <7825, but will in expectation take a very long time for largeish n. TIO starts timing out around n=90.

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1
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Charcoal, 74 bytes

ＮθＦθＦιＦκＦ⁼Ｘ⊕ι²ΣＸ⊕⟦κλ⟧²⊞υ⊕⟦ικλ⟧≔⁰ηＷ¬ⅉ«≔Ｅυ§κ÷ηＸ³λζ≦⊕η≔Ｘζ²ε¿¬⊙ε⊙ε№ε⁺κμＩ⁻Ｅθ⊕κζ

Try it online! Well, for n<50, otherwise it gets too slow. Link is to verbose version of code. Based on @JonathanAllen's answer. Explanation:

Ｎθ

Input n.

ＦθＦιＦκ

Loop through all potential Pythagorean triples.

Ｆ⁼Ｘ⊕ι²ΣＸ⊕⟦κλ⟧²

If this is indeed a triple,

⊞υ⊕⟦ικλ⟧

then push it to the empty list.

≔⁰η

Start iterating through the ways of picking one element of each triple.

Ｗ¬ⅉ«

Repeat until output has been generated.

≔Ｅυ§κ÷ηＸ³λζ

Pick one element from each triple.

≦⊕η

Increment the loop counter.

≔Ｘζ²ε

Square the elements.

¿¬⊙ε⊙ε№ε⁺κμ

Check for Pythagorean triples.

Ｉ⁻Ｅθ⊕κζ

If none, then output one of the sets.

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